Testing Genetic Covariance And Correlation

Why Test Genetic Covariance?

An estimated genetic covariance describes how additive genetic effects on two traits vary together in a population. A nonzero estimate is not, by itself, evidence that the population covariance differs from zero: estimates vary across samples, and uncertainty can be substantial. A formal test asks whether the observed data provide evidence against a specified zero-covariance model, while accounting for the estimation of other model parameters.

Let \(G_1\) and \(G_2\) denote additive genetic effects for two traits. Write their covariance matrix as \(\boldsymbol{\Sigma}_g\), with genetic variances \(\sigma^2_{g,1}\) and \(\sigma^2_{g,2}\), and cross-trait genetic covariance \(\sigma_{g,12}\). When both genetic variances are positive, the genetic correlation is \(\rho_g=\sigma_{g,12}/(\sigma_{g,1}\sigma_{g,2})\). Thus zero covariance and zero correlation are equivalent under that condition. This tutorial concerns testing that null, not interpreting associations as causal effects.

Definition: For a bivariate genetic covariance model with positive genetic variances, the test of zero cross-trait genetic covariance is \(H_0:\sigma_{g,12}=0\) against \(H_a:\sigma_{g,12}\ne 0\). Equivalently, it tests \(H_0:\rho_g=0\) against \(H_a:\rho_g\ne 0\). The alternative is two-sided unless a directional alternative is specified in advance.

The covariance matrix must be positive semidefinite. For two traits this requires \(\sigma^2_{g,1}\ge 0\), \(\sigma^2_{g,2}\ge 0\), and \(\sigma_{g,12}^2\le \sigma^2_{g,1}\sigma^2_{g,2}\). If both variances are strictly positive, zero covariance is an interior point of the allowed range for the covariance. If a variance is zero, however, the covariance must also be zero; the usual equivalence with a correlation and ordinary one-degree-of-freedom calibration can then fail.

Likelihood-Ratio, Score, and Wald Tests

Let \(\psi=\sigma_{g,12}\) be the parameter being tested, and let \(\boldsymbol{\eta}\) collect nuisance parameters, such as the two genetic variances, residual covariance parameters, and other model parameters. A test must compare models with the same data and the same treatment of other parameters. For likelihood-based tests, the null model fixes \(\psi=0\), while the alternative estimates \(\psi\) subject to the covariance-matrix constraints.

Likelihood-ratio test: Let \(\ell_0\) be the maximized log likelihood under \(H_0\), and \(\ell_1\) the maximized log likelihood under \(H_a\). The statistic is \(T_{\mathrm{LR}}=2(\ell_1-\ell_0)\). Under regularity conditions and \(H_0\), it is approximately distributed as \(\chi^2_1\), a chi-square distribution with one degree of freedom.

The likelihood-ratio test measures how much better the best-fitting alternative model fits than the best-fitting null model. The one-degree-of-freedom reference is justified when the null is an interior restriction on one identifiable parameter, the model is correctly specified, and the usual likelihood regularity conditions hold. For restricted maximum likelihood (REML), compare restricted likelihoods only when the fixed-effects structure is the same in the two models.

Score test: Fit the model under \(H_0\), then evaluate the score for \(\psi\) at that constrained fit. With nuisance parameters estimated under the null, let \(U_{\psi\cdot\eta}\) be the nuisance-adjusted score and \(I_{\psi\psi\cdot\eta}\) its efficient information. The statistic is \(T_{\mathrm{Score}}=U_{\psi\cdot\eta}^2/I_{\psi\psi\cdot\eta}\), approximately \(\chi^2_1\) under regularity conditions.

In a regular likelihood model, the nuisance adjustment accounts for the fact that \(\boldsymbol{\eta}\) is estimated rather than known. In terms of the score vector and information matrix evaluated under the null, it uses \(U_{\psi\cdot\eta}=U_\psi-I_{\psi\eta}I_{\eta\eta}^{-1}U_\eta\) and \(I_{\psi\psi\cdot\eta}=I_{\psi\psi}-I_{\psi\eta}I_{\eta\eta}^{-1}I_{\eta\psi}\). This test can be convenient when fitting the alternative model is difficult; it requires the null fit and the relevant score and information calculations.

Wald test: Let \(\hat{\psi}\) be the estimate under the alternative and \(\widehat{\mathrm{Var}}(\hat{\psi})\) its estimated variance. The statistic is \(T_{\mathrm{Wald}}=(\hat{\psi}/SE(\hat{\psi}))^2=\hat{\psi}^{\,2}/\widehat{\mathrm{Var}}(\hat{\psi})\), where \(SE(\hat{\psi})=\sqrt{\widehat{\mathrm{Var}}(\hat{\psi})}\). Under regularity conditions, it is approximately \(\chi^2_1\) under \(H_0\).

The Wald test asks how many estimated standard errors separate the estimate from zero, and squares that standardized distance. It is simple to compute from an estimate and its standard error, but it can behave poorly when the likelihood is asymmetric, the estimate is near a constraint, or the standard error is unstable.

Worked Tests

Worked Example: Likelihood-Ratio Test

Suppose a hypothetical bivariate model has maximized log likelihood \(\ell_0=-102.4\) under zero genetic covariance and \(\ell_1=-100.5\) when the covariance is estimated. Assume the genetic variances are positive, the fitted alternative is within the positive-semidefinite constraint, and the model is regular enough for the asymptotic reference distribution.

The statistic is

T_{\mathrm{LR}}=2\{\ell_1-\ell_0\} =2\{-100.5-(-102.4)\} =2(1.9)=3.8.

Using a \(\chi^2_1\) reference distribution gives a two-sided test p-value of approximately \(0.0513\). At \(\alpha=0.05\), this result does not reject \(H_0\); it is close to the threshold, but is not below it. The p-value is the probability, assuming the null model and the calibration assumptions, of a statistic at least as large as 3.8. It is not the probability that the null hypothesis is true.

Worked Example: Wald Test and the Standard-Error Scale

In a second hypothetical analysis, the estimated genetic covariance is \(\hat{\psi}=0.240\) trait-units squared, with \(SE(\hat{\psi})=0.120\) trait-units squared. Suppose the genetic variance estimates are \(0.64\) and \(1.00\) in the corresponding squared trait units. The covariance estimate is feasible because \(0.240^2=0.0576\) is less than \(0.64(1.00)=0.64\), and both estimated variances are positive.

First standardize the estimate, then square:

T_{\mathrm{Wald}} =\left(\frac{0.240}{0.120}\right)^2 =2^2=4.00.

Equivalently, the estimated variance is \(0.120^2=0.0144\), so \(0.240^2/0.0144=4.00\). Against \(\chi^2_1\), the p-value is approximately \(0.0455\), which is below \(0.05\). Under this test's assumptions, the data provide evidence against zero genetic covariance. The sign of the estimate is positive, but the squared statistic tests departures in either direction.

Worked Example: Score Test at the Null Fit

Suppose a hypothetical model fitted under \(H_0\) gives a nuisance-adjusted score \(U_{\psi\cdot\eta}=1.80\) and efficient information \(I_{\psi\psi\cdot\eta}=9.00\), in mutually compatible parameter units. Assume the null point is interior and regular, with positive genetic variances.

The score statistic is

T_{\mathrm{Score}} =\frac{(1.80)^2}{9.00} =\frac{3.24}{9.00} =0.360.

The \(\chi^2_1\) upper-tail p-value is approximately \(0.5485\). This score test does not reject zero covariance at the 0.05 level. It uses the direction and curvature of the likelihood at the constrained null fit; it does not require the alternative estimate used in the Wald test.

Why the Three Tests Can Differ

Likelihood-ratio, score, and Wald tests are asymptotically equivalent under regularity conditions, but they need not give the same value in a finite sample. The likelihood-ratio test compares fitted likelihoods, the score test measures the local slope at the null fit, and the Wald test measures the alternative estimate relative to its estimated standard error. When the likelihood is approximately quadratic and the estimate is near the null, these views are similar. Skewness, weak information, or proximity to a parameter boundary can make them diverge.

A short local argument explains the connection. In a regular one-parameter problem with nuisance parameters appropriately accounted for, approximate the log likelihood near the null by a quadratic:

\ell(\psi)-\ell(0)\ \approx\ U\psi-\frac{1}{2}I\psi^2,

where \(U\) is the efficient score and \(I\) the efficient information at the null. Maximizing this quadratic gives \(\hat{\psi}\approx U/I\). The likelihood gain is then approximately \(U^2/(2I)\), so twice that gain is \(U^2/I\), the score statistic. The estimated variance of \(\hat{\psi}\) is approximately \(1/I\), making the Wald statistic \(\hat{\psi}^{\,2}/(1/I)\approx U^2/I\) as well. This derivation depends on the local quadratic approximation and does not establish exact finite-sample equality.

Constraints and Small-Sample Calibration

The covariance constraint deserves explicit attention. With positive genetic variances, \(\sigma_{g,12}=0\) is an interior point because small positive and negative covariances are both allowed. In that regular case, the usual one-degree-of-freedom chi-square approximation is appropriate asymptotically. By contrast, testing whether a variance component is zero is a boundary problem: variances cannot be negative. Its null distribution may not be the ordinary \(\chi^2_1\) distribution. A boundary result for a variance test should not automatically be transferred to a zero-covariance test, nor should the covariance constraint be ignored when the fitted variances are near zero or the fitted covariance is near its permitted limit.

In small samples, even an interior null may have a poor chi-square approximation, especially when variance components are weakly identified or the likelihood is markedly non-quadratic. One option is a parametric bootstrap under the fitted null: simulate data from the null model, refit the null and alternative to each simulated data set using the same constraints and fitting procedure, and compare each simulated statistic with the observed one. This estimates the null distribution for the analysis at hand. The model and simulation scheme still need to be credible; bootstrap calibration does not repair misspecification.

Worked Example: Null Parametric Bootstrap

Consider a hypothetical small-sample analysis whose observed likelihood-ratio statistic is \(5.10\). The fitted alternative covariance is close to its permitted upper limit, so the analyst does not rely solely on the asymptotic \(\chi^2_1\) approximation. They simulate \(999\) data sets under the fitted zero-covariance model, refit both models to every simulated data set with the same positive-semidefinite constraints, and find that \(73\) simulated statistics are at least \(5.10\).

Using the common plus-one correction, the bootstrap p-value is

\hat{p}_{\mathrm{boot}} =\frac{73+1}{999+1} =\frac{74}{1000} =0.074.

At \(\alpha=0.05\), this bootstrap-calibrated test does not reject the zero-covariance null. The conclusion differs from what an unexamined asymptotic approximation might suggest, illustrating why constraints and sample size matter. The bootstrap p-value is conditional on the fitted null model and the chosen simulation and refitting procedure.

Choosing and Reporting a Test

For a well-behaved model with adequate information and positive genetic variances, all three tests can be reasonable. The likelihood-ratio test is often a natural choice when both models can be fitted reliably. The score test can be useful when the null fit is straightforward but repeated alternative fitting is costly. The Wald test is convenient when a stable alternative estimate and standard error are available, but should be treated cautiously near constraints. In small or irregular samples, consider null parametric-bootstrap calibration and report how it was performed.

A clear report names the tested parameter, the alternative direction, the statistic and its calibration, the p-value, and the estimated covariance or correlation with units or a dimensionless scale as appropriate. For example, report a covariance in the product of the two trait units; report a correlation without units. Keep estimation and testing distinct: a small p-value is evidence against the specified null under the model, not proof of a causal relation or a measure of practical importance.

Common Pitfall

Miscomputing the Wald statistic. Dividing the squared estimate by its standard error and then squaring again is incorrect. The correct formula is \(T_{\mathrm{Wald}}=(\hat{\psi}/SE(\hat{\psi}))^2=\hat{\psi}^{\,2}/\widehat{\mathrm{Var}}(\hat{\psi})\). With estimate \(0.240\) and standard error \(0.120\), it is \((0.240/0.120)^2=4.00\).
Assuming every covariance-component null is a boundary null. Zero cross-trait covariance is interior when both variances are positive; a zero variance component is different. Check the parameter constraints and the null point before selecting a reference distribution.
Treating zero covariance and zero correlation as interchangeable without conditions. They are equivalent only when both genetic variances are positive. If a genetic variance is zero, correlation is not defined in the usual way.
Reading a p-value as the probability of the null. A p-value is calculated under the null and describes the tail probability of a test statistic at least as extreme as observed; it does not assign a probability to the null hypothesis.

Key takeaway: For a regular one-parameter test of zero genetic covariance with positive genetic variances, likelihood-ratio, score, and Wald statistics have an asymptotic \(\chi^2_1\) reference distribution. Their finite-sample behavior can differ, so check constraints and consider null-based bootstrap calibration when the sample or fit is problematic.

Check Your Understanding

Answer each question using the model assumptions and test definitions above.

When are zero genetic covariance and zero genetic correlation equivalent hypotheses?
A null and alternative model have maximized log likelihoods \(-80.2\) and \(-78.7\). Compute the likelihood-ratio statistic and state its usual asymptotic degrees of freedom under regularity conditions.
An estimate is \(0.18\) with standard error \(0.09\). Compute the Wald statistic, showing the correct order of operations.
Why can a zero genetic variance require different calibration from a zero cross-trait genetic covariance?
In a null parametric bootstrap with \(1{,}999\) simulations, \(119\) statistics are at least as large as the observed statistic. Compute the plus-one corrected bootstrap p-value.